Q.The $E^\circ(M^{2+}/M)$ value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high $\Delta_a H^\circ$ and low $\Delta_{hyd} H^\circ$)

The key idea is that the standard reduction potential $E^\circ$ reflects the net energy change of the overall process: sublimation, ionization, and hydration. For copper, the unusually high enthalpy of atomization ($\Delta_a H^\circ$) and relatively low hydration enthalpy ($\Delta_{hyd} H^\circ$) make the overall conversion of $M(s)$ to $M^{2+}(aq)$ less favourable than for other metals.

Reasoning:

1. The reduction potential $E^\circ(M^{2+}/M)$ is linked to the Gibbs energy change for $M^{2+}(aq) + 2e^- \rightarrow M(s)$. A more positive $E^\circ$ means the reverse (oxidation) is less spontaneous.
2. For copper, the high $\Delta_a H^\circ$ (strong metallic bonding) means a lot of energy is needed to break the metal lattice into gaseous atoms. …

The standard reduction potential $E^\circ$ for a metal ion $M^{2+}$ is a measure of how easily the ion gains electrons to become the metal. A positive value means the reduction is spontaneous relative to the $H^+/H_2$ couple. But why is copper’s value positive, while many other metals (like zinc, iron) have negative values?

The answer lies in the energy changes that occur when a solid metal is converted to its aqueous ions — and then back again. The key is to think of the reduction process in reverse: the oxidation of the metal to its ions.

The Born–Haber cycle for a metal electrode

For the half‑reaction

the overall enthalpy change can be broken into three steps:

1. Atomisation — converting the solid metal into gaseous atoms:

2. Ionisation — removing two electrons from the gaseous atom:

3. Hydration — dissolving the gaseous ion in water:

The total enthalpy change for the oxidation is

The reduction potential is related to the reverse of this process. A more positive $E^\circ$ means the reduction $M^{2+}(aq) + 2e^- \to M(s)$ is more favourable — which corresponds to a less favourable oxidation (i.e., a larger positive $\Delta H_{ox}$).

Why copper stands out

For most transition metals, $\Delta_a H^\circ$ is moderate and $\Delta_{hyd} H^\circ$ is highly negative (strong ion–water attraction), making $\Delta H_{ox}$ negative overall — so oxidation is easy, and $E^\circ$ is negative.

Copper is different:

• High $\Delta_a H^\circ$ — Copper has strong metallic bonding (due to its filled $d^{10}$ configuration and efficient packing), so it takes a lot of energy to break the metal into atoms.
• Comparatively low $\Delta_{hyd} H^\circ$ — the energy released when $Cu^{2+}$ is hydrated, though substantial, is not large enough to pay back copper's unusually high atomisation-plus-ionisation cost. This is exactly the balance the question's own hint points to: high $\Delta_a H^\circ$, low $\Delta_{hyd} H^\circ$.

These two factors together make $\Delta H_{ox}$ less negative (or even positive) for copper. That means the oxidation $Cu(s) \to Cu^{2+}(aq) + 2e^-$ is less spontaneous — and conversely, the reduction $Cu^{2+}(aq) + 2e^- \to Cu(s)$ is more spontaneous, giving a positive $E^\circ$.

Step‑by‑step reasoning

1. Recall the definition — $E^\circ$ measures the tendency of $M^{2+}(aq)$ to gain electrons. A positive value means the reduction is favoured over the $H^+/H_2$ couple.

2. Break the reduction into its reverse (oxidation) — The easier it is to oxidise $M(s)$ to $M^{2+}(aq)$, the less positive (more negative) $E^\circ$ will be. …

Method: Born–Haber Cycle Reasoning for Standard Reduction Potential

This method uses the thermodynamic cycle approach to explain why a metal’s reduction potential is positive or negative by breaking the overall process into enthalpy steps.

Steps

1. Write the half-reaction For copper:

2. Reverse the process to think in terms of formation of the ion Consider the reverse (oxidation):

3. Break the oxidation into two conceptual steps
   • Step A: Convert solid metal to gaseous metal ions (atomization + ionization)

 Here $\Delta_a H^\circ$ is the **atomization enthalpy** (energy to separate solid into gaseous atoms).

• Step B: Hydrate the gaseous ions

 $\Delta_{hyd} H^\circ$ is the **hydration enthalpy** (energy released when ions are surrounded by water).

4. Relate to $E^\circ$

The overall enthalpy change for oxidation is:

A more positive $E^\circ$ (easier reduction) means the reverse reaction (reduction) is more spontaneous — i.e., the oxidation is less favourable (more endothermic).

5. Apply to copper
   • Copper has a high $\Delta_a H^\circ$ (strong metallic bonding, high melting point). …

Here is a breakdown of the common mistakes students make when answering this question, along with the correct conceptual approach.

The Core Concept (The "Why")

The question asks why Copper has a positive $E^\circ$ value. A positive value means $Cu^{2+}$ ions have a low tendency to get reduced (gain electrons) compared to the standard hydrogen electrode.

The hint points to two key energy terms:

1. High $\Delta_a H^\circ$ (Atomisation Enthalpy): This is the energy needed to turn solid Copper into gaseous atoms. Copper has a high value because of strong metallic bonding.
2. Low $\Delta_{hyd} H^\circ$ (Hydration Enthalpy): This is the energy released when gaseous $Cu^{2+}$ ions are surrounded by water. For copper this release, though substantial, is not large enough to compensate the unusually high atomisation-plus-ionisation input — which is the sense in which the hint calls it "low".

The "Why" in one line: The overall energy change for the process $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$ is less favourable (less exothermic or more endothermic) than for other metals, making the reverse reaction (reduction) comparatively favourable, hence the positive $E^\circ$.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Positive $E^\circ$" with "Easily Reduced"

• The Error: Students think a positive $E^\circ$ means the metal is easily reduced. They then say "Copper is easily reduced, so it has a high tendency to gain electrons."
• Why it's wrong: A positive $E^\circ$ means the reduction is less spontaneous compared to the standard hydrogen electrode. It means the metal ion is a weak oxidising agent.
• How to Avoid: Always remember the Electrochemical Series.
  • More positive $E^\circ$ = Stronger oxidising agent (easier to reduce the ion, harder to oxidise the metal). Example: $F_2$ (+2.87V).
  • More negative $E^\circ$ = Stronger reducing agent (easier to oxidise the metal, harder to reduce the ion). Example: $Li$ (-3.04V).
  • Copper (+0.34V) is a weak reducing agent (hard to oxidise) and a weak oxidising agent (hard to reduce). The positive value tells you the metal is noble (resists oxidation).

Mistake 2: Ignoring the Hint & Using the Wrong Energy Terms

• The Error: Students try to explain the positive $E^\circ$ using only Ionisation Enthalpy ($\Delta_i H$) or Electron Gain Enthalpy ($\Delta_{eg} H$). They say "Copper has high ionisation enthalpy, so it's hard to remove electrons, hence positive $E^\circ$."
• Why it's wrong: While ionisation enthalpy is part of the story, the hint specifically asks you to consider Atomisation and Hydration. The overall energy for $M(s) \rightarrow M^{2+}(aq)$ is:

For Copper, the **high $\Delta_a H^\circ$** (strong metallic bonds) and **low $\Delta_{hyd} H^\circ$** (less energy released on hydration) make the overall process **less exothermic** (or more endothermic) than for other metals. This makes the reverse reduction reaction less spontaneous, giving a positive $E^\circ$.

• How to Avoid: When a question gives a hint, use it. Write down the three-step Born-Haber cycle for the formation of aqueous ions:
  1. $M(s) \xrightarrow{\Delta_a H^\circ} M(g)$
  2. $M(g) \xrightarrow{\Delta_i H^\circ} M^{2+}(g) + 2e^-$
  3. $M^{2+}(g) \xrightarrow{\Delta_{hyd} H^\circ} M^{2+}(aq)$ Then explain how the high $\Delta_a H^\circ$ and low $\Delta_{hyd} H^\circ$ for Copper make the overall process unfavourable.

Mistake 3: Misreading the Hint's "Low $\Delta_{hyd} H^\circ$" as a Comparison Across Elements

• The Error: Students write "Copper's hydration enthalpy is smaller (less negative) than that of other $M^{2+}$ ions like $Zn^{2+}$", treating the hint's 'low' as an element-by-element comparison.
• Why it's wrong: Table 4.4 gives $\Delta_{hyd}H^\circ(Cu^{2+}) = -2121$ kJ mol$^{-1}$ — actually more negative than $Zn^{2+}$ ($-2059$) and among the most negative of the whole series. The hint's 'low' means not large enough: the energy released on hydration does not pay back copper's unusually high atomisation + ionisation input.
• How to Avoid: Be precise with the energy bookkeeping.
  • Hydration is the only energy-releasing step in $Cu(s) \rightarrow Cu^{2+}(aq)$; atomisation and ionisation both cost energy.
  • For most 3d metals the hydration release outweighs a modest input, so their $E^\circ$ is negative.
  • For copper, the input (high $\Delta_aH^\circ$ plus the very high $\Delta_iH_1 + \Delta_iH_2$) exceeds the hydration payback, so oxidation is unfavourable and $E^\circ$ is positive.

Mistake 4: Forgetting the Standard Hydrogen Electrode (SHE)

• The Error: Students explain the positive value in isolation, without comparing it to the SHE. …